Let t denote the greatest integer < t. If for some


Let $[t]$ denote the greatest integer $\leq t$. If for some

$\lambda \in \mathrm{R}-\{0,1\}, \lim _{\mathrm{x} \rightarrow 0}\left|\frac{1-\mathrm{x}+|\mathrm{x}|}{\lambda-\mathrm{x}+[\mathrm{x}]}\right|=\mathrm{L}$, then $\mathrm{L}$ is

equal to :

  1. 1

  2. 2

  3. $\frac{1}{2}$

  4. 0

Correct Option: , 2


$\mathrm{LHL}: \lim _{\mathrm{x} \rightarrow 0^{-}}\left|\frac{1-\mathrm{x}-\mathrm{x}}{\lambda-\mathrm{x}-1}\right|=\left|\frac{1}{\lambda-1}\right|$

$\mathrm{RHL}: \lim _{x \rightarrow 0^{+}}\left|\frac{1-x+x}{\lambda-x+1}\right|=\left|\frac{1}{\lambda}\right|$

For existence of limit


$\Rightarrow \frac{1}{|\lambda-1|}=\frac{1}{|\lambda|} \Rightarrow \lambda=\frac{1}{2}$

$\therefore \quad \mathrm{L}=\frac{1}{|\lambda|}=2$

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